Ch 14 power point

Chapter 14
Periodic Trends
CP Chemistry
Fall 2013
Chapter 14: Chemical Periodicity
• 14.1 Objectives
– Explain why you can infer the properties of an
element based on those of other elements in the
periodic table.
– Use electron configuration to classify the elements
Noble gases
Representative elements
Transition metals
Inner transition metals
The Periodic Table
• Russian chemist Dmitri Mendeleev, in 1871, created a successful
organizational scheme for the periodic table.
• The periodic table is useful for understanding and predicting the
properties of the elements.
– If you know the chemical and physical properties of one
element of a group (vertical column) you can forecast the
properties of the other elements in that column too.
• The electron plays the most significant role (more than protons
or neutrons) in determining the physical and chemical properties
of an element.
– This is because it is on the “outside” of the atom
• Therefore there is a relationship between the electron
configuration and the placement of elements in the table.
See how you add one electron to the outer shell as you move
from left to right?
Shell 2 contains the 2s and 2p orbitals for a total of 8 electrons.
Which element has 8 electrons in the 2nd shell?
See how you add a new outer shell
every time you start a new row of
the periodic table?
Period 1 has 1 shell
Period 2 has 2 shells
Period 3 has 3 shells and so forth…
Then if you compare Be to Li, see
how it has one more electron in the
outer shell?
Same thing if you compare
Potassium and Calcium. See how
one electron got added to the outer
But what happens with Scandium?
Why aren’t there 3 e- in the outer
shell? There is an extra one but
where is it and why??
Noble Gases
• These are elements where the outermost s
and p sublevels are filled up (2 electrons in
each “square” (orbital), all squares in Aufbau
diagram are full.
• The noble gases are called Group 0.
• These elements are also called inert gases
because they are very unreactive with other
– Their outer sublevels are “full” so they don’t want
to gain or lose any more electrons.
Noble gases (continued)
• Look at the periodic table on page 393. Note the
top right hand corner shows the electrons in
each energy level.
• He 1s2 (total 2)
• Ne 1s22s22px22py22pz2 or 1s2 2s22p6 (total 10)
• Ar 1s2 2s22p6 3s23p6 (total 18)
• Kr
1s2 2s22p6 3s23p63d10 4s24p6 (total 36)
Representative Elements
• For these elements, the outermost s or p level is partially filled.
• The representative elements are usually called Group A elements.
– Sometimes the noble gases are also included in Group A.
• Group 1A (1st column on far left) elements are called alkali metals
• Group 2A (2nd column) elements are called alkaline earth metals.
• Group 7A (2nd to the right column) elements are called halogens.
• For any representative element, the number (2A, 7A) equals the
number of electrons in the outermost energy level.
• Example Group 1A: One electron in outer sublevel (n=2,3,4)
• Li n=2
1s2 2s1
• Na n=3
1s2 2s2 2p6 3s1
• K n=4
1s2 2s2 2p6 3s2 3p6 4s1
Representative Elements (cont.)
• Let’s try Group 4A. It should have 4 e- in outer
– C 1s2 2s2 2p2 (2 e- in 2s, 2 e- in 2p, total=4)
– Si 1s2 2s2 2p6 3s2 3p2 (4 e- in sublevel 3)
– Ge 1s2 2s2 2p6 3s2 3p6 4s2 4p2 (4 e- in sublevel 4)
• How many e- in outer sublevel/shell for
Mg ?
I ?
The transition metals
• These are metallic elements in which the
outermost s sublevel and nearby d sublevel
contains electrons.
• The transition elements, called the Group B
elements, are characterized by the addition of
electrons to the d-orbitals.
• Example: Sc in column 3B
– 1s2 2s2 2p6 3s2 3p6 3d1 4s2
• Did you notice that the d level you are filling is
one less than the last s level you filled? (3 vs. 4)
The inner transition metals
• These are metallic elements in which the
outermost s sublevel and the nearby f sublevel
generally contain electrons.
• The inner transition metals are characterized
by the filling of the f-sublevels.
• When you are filling f-sublevels, you will be
working with a number 2 less than the last slevel filled. For example, when you finish with
6s2 then you will be filling the 4f level
Note that the d-orbitals are associated with the transition metals.
The f-orbitals are associated with the inner transition metals.
Note how the d-levels are one less than the matching s-levels.
The f-levels are two less than the matching s-levels.
• If you consider electron configurations and the position of the
elements in the periodic table, a new pattern can be seen.
The periodic table can be divided into blocks that correspond
to which sublevels are filled with electrons.
Outer shell electrons
• s-block – how many
outer electrons?
– Is He included?
– How many
electrons does it
• p-block – what groups
does it include?
– Does it include
Electron configurations
• Electron configurations can be found by reading the periodic
table like you read a book, from top to bottom, left to right.
• Each period (line you are reading) corresponds to one
principal energy level (n=1,2,3)
• You can find the number of e- in each sublevel by counting
from left to right.
• For transition elements, e- are added to a d sublevel with
principal energy level one less than the period number.
– Example: if you are in row 4, you will be adding e- to level
• For inner transition elements, it’s the same except for f
sublevels it will be two less than the principal energy level.
– Example: If you are in row 6, adding e- to level 4f
Sample problem 14-1
• Write the electron configuration for N and Ni.
a. N has 7 protons (atomic number) and therefore 7
e-. It is in row 2 so it’s principal energy level = 2.
Start filling until you get to 7 total: 1s2 2s2 2p3
N is the 3rd element over in the p-block so it
makes sense that it should have 2p3
b. Ni has 28 protons and 28 e-. Its in row 4, so n=4
1s2 2s2 2p6 3s2 3p6 3d8 4s2 total superscripts = 28
(recall you are filling d-level 3 after s-level 4, so dlevel 3 isn’t full yet…)
Section 14.2 – Periodic Trends
• Objectives –
– Interpret group trends in
Atomic radii
Ionic radii
Ionization energies
– Interpret period trends in the same four items
listed above.
Atomic radii
• To measure atomic
size (atomic radii),
x-ray diffraction can
be used to measure
the distance between
nuclei of a diatomic
molecule like H2, Cl2
or O2.
• 1 pm = 1 picometer
= 1x10-12m
What trends in atomic radii do you see ?
• Look at the top right corner for size in pm (picometers).
• Look at size of atom drawn in top left corner.
• Ignore ionic radii (bottom of cells) for now.
Group Trends – Atomic Radius
• Atomic size generally increases as you move down a
group in the periodic table.
• As you move down, there are more e- being added to
successively higher energy levels, so that the outer
orbital is further out as you move downwards.
• The nuclear charge (# protons) is also increasing at the
same rate. The inner orbitals filled with e- have a
shielding effect.
• You might expect that the + charge in the nucleus
would pull e- towards the center and shrink the size of
the atom, but that isn’t what happens.
• The enlarging effect of the greater distance of the
outer electrons overcomes the shrinking effect caused
by the increased + charge in the nucleus.
Atomic vs. Ionic radii
• Note how the atomic radius (grey
bars) increase as you go down the
group (from atomic radius: Li < Na
< K < Rb < Cs
• Now look at the blue bars. These
are positive ions. That means that
one outer shell electron has been
stolen from the atom. Why do you
think that makes the atomic radius
smaller (why are the blue bars
shorter than the grey bars)?
• Look at the bottom chart for the
halogens. If you now add an e-, the
ionic radius gets larger, why?
Periodic Trends
You may see the attraction
between oppositely charged
particles, such as an electron
and a proton, called
Coulombic attraction in your
• Atomic size generally decreases as you move from left to right
across a period.
• As you go across a period, the principal energy level is constant. As
you go across, one proton and one e- are being added per square.
• The effect of the added protons to the nucleus is to attract the
electrons closer to the nucleus, shrinking atomic size.
• If you plot an atomic radius (y-axis) vs. atomic number (x-axis) you
get the graph here.
Quick Quiz 1
– Why is atomic radius going down from Li to Ne?
1s2 2s1
1s2 2s2
1s2 2s2 2p1
1s2 2s2 2p2
1s2 2s2 2p3
1s2 2s2 2p4
1s2 2s2 2p5
1s2 2s2 2p6
1s2 2s2 2p6 3s1
1 outer e2 outer e3 outer e4 outer e5 outer e6 outer e7 outer e8 outer e-, full p-shell
Why is Na radius larger??
• Quick quiz #2 –
• Can anyone explain why the size
jumps up with Na?
• Neon had 10 protons, Na has 11
protons, wouldn’t the addition of
one more proton make the atom
smaller due to attraction from the
• It is bigger because now that all the
2p orbitals are full, the last e- went in
the 3s shell (n=3 principal number),
which is further out from the
nucleus, making the atom larger.
• QQ #3: Why is period 4’s downward trend more gradual
than period 2’s trend?
• There is a shielding effect that occurs for the higher
electron counts. The nucleus does have a lot of + charge,
but the numerous inner e- interfere and shield the effect
of the nucleus pulling in the outer e- as much.
Trends in Atomic Radii Summary
Memorize this page for the test!
Trends in Ionization Energy
• When a neutral atom gains or loses an e-, it
becomes an ion. (Note it doesn’t lose/gain any
protons or neutrons).
• The + nucleus exerts an attractive force on all its
e-. It doesn’t want to lose any e-, so to remove an
e- requires energy to overcome this force. This
amount of energy (different for every element) is
called the ionization energy.
• Na (g)
Na+(g) + e• The energy required to remove the first e- is
called the first ionization energy. If a second e- is
removed, the energy required to do that is the
second ionization energy.
• Mg
Mg2+ + 2e-
Look at oxygen
• What are its 1st, 2nd and 3rd
ionization energies?
1314, 3391, 5301 kJ/mol
• What do you notice about the
Each electron gets harder to
Look at Na
• Why is there a dashed line
between the 1st and 2nd
ionization energies?
Na 1s2 2s2 2p6 3s1
1st ioniz. energy: 495
2nd ioniz. energy: 4565
First e- takes partially filled 3s1 e-,
2nd e- breaks into full p-shell,
doesn’t want to give up that e-.
Look at Li >> Ne:
• What trend do you notice in 1st
ionization energy?
• Ionization energy increases
• Can you hypothesize why that
might be?
• From B to Ne, the closer the pshell is to being full, the more it
wants to stay full and the more
difficult it is to steal an e-.
• Why is the first ionization energy
for B lower than that for Be?
That doesn’t follow the trend.
• Be has a full 2s2 shell, doesn’t
want to have that e- stolen. B
has only 1 e- in p-shell, doesn’t
mind losing it.
Why do you think the first
ionization energy decreases as
you move down a group? For
Li 520 kJ/mol
Na 495 kJ/mol
K 419 kJ/mol
This is because the size of the
atoms increases as you go
down. As there are more e-,
they fill orbitals/shells that are
further out from the nucleus,
which means the attractive
force from the nucleus is less
because they are farther away.
Therefore they are easier to
Trends in Ionization Energy
Notice how those
noble gases (He, Ne,
Ar…) are not willing
to give up their fullshell electron which
means they have a
high first ionization
For the representative elements, first ionization energy
increases going from left to right. As you move across, it adds
a p+ and an e- with each square. A greater charged nucleus =
a greater attraction = less willing to give up an e-.
Ionization Energy Trends
Memorize this chart for the test!
QQ: Which group 6A element has the highest first
ionization energy? Which element in period 2 has the
highest ionization energy?
Naming Ions
• Before we start discussing the trends in ionic
radii, let’s first define a few things:
• Cation – a positive ion. That means that an
electron has been stolen from the neutral atom
and it has 1 more proton than electron.
• Anion – a negative ion. It has a negative charge
because it has stolen an electron from some
other atom, meaning it has one more e- than it
does protons, giving it a negative charge.
• Refer to pages 135-136 in your book for this
content (It is in Ch. 6, which we haven’t read yet.)
Naming Cations – positive ions
• Let’s say sodium, Na, has 11 protons and only
10 electrons (one has been stolen). How do
we write that ion? Na+ called “Sodium ion”
• What if we had Ca, with 20 protons and only
18 electrons (the two outer shell e- have been
stolen). How do we write that ion? Ca2+ and
we call that “Calcium ion”
Naming Metal Cations
• This actually comes from Ch. 6 page 144, but let’s have a
preview now:
• Cu+ ion is called a Copper(I) ion.
• Cu2+ ion is called a Copper(II) ion.
• Sn2+ ion is called a Tin(II) ion. You say “Tin two ion”
• Sn4+ ion is called a Tin(IV) ion. You say “Tin four ion”, etc.
• Mn2+ ion is called what?
• Mn3+ ion is called what?
• Note that many metals can have more than one ionic
charge (although a particular one may be most common).
• Note that you don’t put any space between the name and
the parenthesis that follow it.
• Need to know this for this test!
Naming Anions – negative ions
• Atoms of nonmetallic elements tend to form
anions because they want to add e- to fill their
shell until it looks like a noble gas.
• Cl- has 17 protons and 18 e-, one extra that it
has stolen. This is called a “chloride anion.”
• Note the –ide ending on anions!
• O2- is the oxide anion that you get when
oxygen, with 8 protons, gains/steals two eand then has 10 e-. The superscript is telling
you that it has two too many electrons relative
to its neutral form.
Why does it change size so much when it becomes an ion?
You are about to find out.
Trends in Ionic Radii/Size
• The atoms of metallic elements have low
ionization energies, which means they form
positive ions easily.
• That means they don’t hold onto their outer every strongly. That’s why metals conduct
electricity, there is a “sea of electrons”
surrounding the metal nuclei and those free ebecome the electric current when they freely
• By contrast, nonmetallic elements readily form
negative ions (they steal e-).
• Can you explain why that Li atom got smaller after one electron
got stolen from it?
• Li is 1s2 2s1 . If it’s outer electron gets stolen, then it’s back to
the 1s orbital, which is closer to the nucleus. The same thing
occurs with Na and K, who each have only one outer e-.
• What about Aluminum when 3 e- get stolen?
• Al has three p electrons in the outer 2p shell. When all three
are stolen, it is also back to it’s 1s orbital, which is smaller.
• Now what about the halogen atoms, F, Cl and Br … Why does
their size get much larger when one electron is added to them?
• Let’s start with Fluorine 1s2 2s2 2p5
• This has 7 outer shell electrons (needs 8 to be full), so it is not
larger because the last e- goes into 3s1, it doesn’t – it goes 2p6.
So why is the – (negative) ion greater in size?
• Now you have one more e- than you do protons, so the force per
e- is now weaker than it is when the number matches, so it
Group Trends - Ions
• Positive ions (cations) are always smaller than
the neutral atom they start out with – why?
• The number of protons is still the same, so the
force per e- is stronger and it pulls it inward.
• In contrast, negative ions (anions) are always
larger than the neutral atom because now the
same proton charge has 1 extra e- to attract,
which makes the average force per e- less.
• See graphic on next page…
Now look again at ionic radii
Periodic Trend for
Ionic Size/Radii
• Going from L to R across a row, there’s a
gradual decrease in the size of positive ions.
• Then, starting with group 5A (Nitrogen), the
negative ions, which are much larger,
gradually decrease in size as you move from L
to R also.
– N3- O2- F- : trend makes sense, each one has fewer
extra electrons so the effect is smaller.
Last but not least, trends in
• The electronegativity of
an atom is the tendency
of an atom to attract
(steal) e- when it is
chemically combined with
another element.
• Relative electronegativity
values are expressed on
the Pauling
electronegativity scale.
• Highest: F 4.0
Lowest: Cs 0.7
• What trends do you see going
across a period L to R? Why?
• Down a group? Why?
• Metallic elements at far left have low electronegativity.
– Remember earlier when we said the e- for metals are
loosely associated with the nucleus like a sea of electrons?
• Nonmetallic elements at far right have high
– Fluorine is the highest at 4.0 – its shell lacks just 1 e- to
have a full 2p6 so it really “wants” that e-. Since F is so
likely to steal an e-, it naturally forms F- ions.
– Cesium, with electronegativity of 0.7 only has one outer
shell e-, and it easily gives it up to form Cs+ ions. Use a
similar argument for Na+ ions.
• Electronegativity values are used to predict the type of
bonding that will occur between 2 atoms. Na-Cl
becomes Na+Cl- ionic bonding, while a C-H bond with
electronegativity values of 2.1 and 2.5 forms a covalent
bond where the e- are shared rather than “stolen”.
Summary Chart –
Need to memorize
QQ#1 : Explain why
atomic radius goes
down as you go L to R.
Shielding is constant
as you go L to R (same
principal quantum
number) but more
protons mean greater
attraction to pull e- in
that shell inwards.
Summary Chart –
Need to memorize
QQ#2 : Explain why
ionization energy goes
up as you go from
L to R.
Group 1 – willing to
give up its one outer
electron easily.
Group 0 – full outer
shell, doesn’t want to
lose any e-
Summary Chart –
Need to memorize
QQ#3 : Explain why
electronegativity increases
as you go from L to R (up
to Group 7).
Group 1 – willing to give
up its one outer electron
Group 7 – Badly wants to
steal an e- to complete its
principal energy level’s
outer shell
Summary Chart –
Need to memorize
QQ#4 : Why is nuclear
charge increasing from L
to R?
Nuclear charge is
increasing because there
is one more proton each
time you move one square
to the right.
Summary Chart –
Need to memorize
QQ#5 : Why does atomic
radius go up as you go
down a group?
Protons are being added
to the nucleus, and
therefore e- are filling
shells farther out from the
nucleus, increasing the
size of the atom.
Summary Chart –
Need to memorize
QQ#6 : Why does
ionization energy go down
as you go down a group?
Because outer e- are
further from the nucleus
and so there is less
attractive force from the
nucleus out there,
meaning it is easier to
steal an e-.

similar documents